What is the integral of (2x)/(x^2-4) ?

The integral of (2x)/(x^2-4) is ln|x^2-4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(x^2-4)

\int \frac{2 x}{x ^{2} - 4} d x

ln x^{2} - 4 + C

Solution steps

\int \frac{2 x}{x ^{2} - 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{x ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 2 \cdot \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 4

= 2 \cdot \frac{1}{2} ln x^{2} - 4

Simplify

2 \cdot \frac{1}{2} ln x^{2} - 4 : ln x^{2} - 4

= ln x^{2} - 4

Add a constant to the solution

= ln x^{2} - 4 + C

\int - cos (x) + 1 d x

x \to 0 lim (\frac{( x ^{2} + 2 x )}{x})

\frac{\partial}{\partial y} (- 2 x y (- 15 - x - y))

\int_{0}^{1} \frac{2}{x ( x + 1 )} d x

y = 4.2 x - x^{2}

What is the integral of (2x)/(x^2-4) ?
The integral of (2x)/(x^2-4) is ln|x^2-4|+C